🔄Update: 螺旋矩阵

矩阵/1. 螺旋矩阵.md

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+# 螺旋矩阵
+
+## 题目
+
+给你一个 m 行 n 列的矩阵 matrix ，请按照 顺时针螺旋顺序 ，返回矩阵中的所有元素。
+
+ 
+
+示例 1：
+
+
+输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]
+输出：[1,2,3,6,9,8,7,4,5]
+示例 2：
+
+
+输入：matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
+输出：[1,2,3,4,8,12,11,10,9,5,6,7]
+ 
+
+提示：
+
+m == matrix.length
+n == matrix[i].length
+1 <= m, n <= 10
+-100 <= matrix[i][j] <= 100
+
+## 思路
+
+- 模拟
+  - 方向
+    - →↓←↑
+  - 边界
+    - `up`
+    - `down`
+    - `left`
+    - `right`
+
+## 代码
+
+```java
+class Solution {
+    public List<Integer> spiralOrder(int[][] matrix) {
+        // Init: up, down, left, right
+        int up = 0;
+        int down = matrix.length - 1;
+        int left = 0;
+        int right = matrix[0].length - 1;
+        // Init: ans
+        List<Integer> ans = new ArrayList<>();
+        int tar = matrix.length * matrix[0].length;
+        // Traverse
+        while (left <= right && up <= down) {
+            for (int i = left; i <= right; i++) {
+                ans.add(matrix[up][i]);
+            }
+            up ++;
+
+            for (int i = up; i <= down; i++) {
+                ans.add(matrix[i][right]);
+            }
+            right --;
+            
+            if (up <= down) {
+                for (int i = right; i >= left; i--) {
+                    ans.add(matrix[down][i]);
+                }
+            }
+            down --;
+            
+            if (left <= right) {
+                for (int i = down; i >= up; i--) {
+                    ans.add(matrix[i][left]);
+                }
+            }
+            left ++;
+
+        }
+
+        return ans;
+    }
+}
+```